C4. Chemical Calculations (+Do Example Q's) (Y10 - Spring 1)

Question 1

🟢 Units and Apparatus for Volume

Answer 1

Chemical Volumes are measured in centimetre cubed or decimetre cubed.
The symbol for the unit is cm^3 or dm^3

Apparatus: A biuret can be used, or a measuring cylinder. (For a solutionary liquid).

Question 2

🟢 Units and Apparatus for Mass

Answer 2

Chemical masses are measured in grams. The symbol for the unit is g.

Apparatus: A ‘Mass Balance’ is used. (For a solid mass)

Question 3

🟢 Units and Apparatus for Amount

Answer 3

Chemical amounts are measured in moles. The symbol for the units of mole is mol.

Apparatus: No specific device/apparatus - it cannot be measured. The amount can’t be measured (can be caluclated from the mass or the volume

Question 4

🟢 What is Avogadro’s Constant

Answer 4

The Avogadro Constant is the number of particles in one mole of particles. It has a value of 6.02 x 10^23

Question 5

🟢 What is a mole?

Answer 5

One mole of a substance contains the same number of particles, atoms, molecules or ions as one mole of any other substance.

The number of atoms, molecules or ions in one mole of a given substance is the Avogadro constant.

The value of the Avogadro constant is 6.02x10^23 per mole.

Question 6

🟢 Explain why relative atomic masses are used rather than actual masses of atoms

Answer 6

Relative atomic masses are used rather than actual masses, because they are all relative to Carbon-12, which has an actual mass of 12.0 for 1 mol of it.

Question 7

🟢 Equation Triangle for Mass, Moles, and Mr (relative molecular mass)/Molar Mass

Answer 7

.              
             /^\
           //     \\ 
         / mass \
       / \_\_\_\_\_\_ \
     /         |        \
    /moles| Mr   \
   /\_\_\_\_\_|\_\_\_\_\_\

Question 8

🟢 How to find if one substance contains more atoms in 1 mol than another substance (e.g Which contains more atoms in 1 mol: 56g of Fe (Iron), or 16g of S (Sulphur))

Answer 8

Mass (g) / Mr = Amount in mols
Amount of Mols x 6.02x10^23 = The Number of Atoms
Then see which substance has a higher amount of atoms.

Fe: 56g / 56 (Mr) = 1 mol = 6.02x10^23
S: 16g / 32 (Mr) = 0.5 mols = 3.01x10^23

This shows that Fe has more atoms in 1 mol than S does.

Question 9

🟢 How you can find the mass of one golf ball, when a dozen golf balls with a mass if 551.16g

Answer 9

Mass of the 12 Golf Balls / The amount of Golf Ball = The Mass if one Golf Ball

551.16 / 12 =45.9

Question 10

🟢 How to Calculate the number of mols in a Substance When you know the Mass and the Substance? (e.g 90.0g of H2O)

Answer 10

Firstly, get the Relative Atomic Mass (Mr) of the substance, then do the Mass / Mr = Mols.

Hydrogen: 1 x 2 = 2
Oxygen: 16
H2O = 2 + 16 =18

90g / 18 = 5 mols

Question 11

🟢 How to Calculate the mass of a Substance When you know the number of mols and the Substance? (e.g 2.5 mols of PH3)

Answer 11

Firstly, get the Relative Atomic Mass (Mr) of the substance, then do the Number of Mols x Mr = Mass

Phosphorus: 31
Hydrogen: 1 x 3 = 3
PH3: 31 + 3 = 34

2.5 mols x 34 = 85g

Question 12

🟢 How to Calculate the Formula Mass of a Substance When you know the number of mols and the mass? (e.g 0.02 mols and a mass of 1.64g)

Answer 12

In this scenario, you just need to remmember the mass number is the Mass / Mols = Relative Formula Mass

0.02 / 1.64 = 82

Therefore, the Relative Formula Mass is 82

Question 13

🟢 How to see if there are more atoms in a certain mass of one substance, than the same mass of another substance (e.g How many more atoms are there in 48g of C compared to 48g of Mg)

Answer 13

Firstly, find out the number of atoms by doing
Mass (g) / Mr = Amount in mols
Amount of Mols x 6.02x10^23 = The Number of Atoms
Then compare the number of atoms

C: 48/12 = 4 mols = 2.408x10^24
Mg: 48/24 = 2 mols = 1.204x10^24

2.408x10^24 - 1.204x10^24 = 1.204x10^24

Question 14

🟢 How to calculate which substance is heavier when you know the substance and the amount of mols (e.g 10 moles of He or 1.5 moles of O2)

Answer 14

Mols x Mr = Mass (g)

He: 10 x 4 = 40g
O2: 1.5 x 32 = 48g

The 1.5 moles of O2 is heavier.

Question 15

🟠 What is Formula + What is Empirical Formula + Molecular Formula

Answer 15

A formula shows us the mole ratio of atoms witnin a substance

The Empirical formula shows us the simplest (whole number) mole ratio of atoms of each element present within a compound.

The Molecular formula shows us the actual (whole number) mole ratio of atoms of each element within a molecule.

Question 16

🟠 How do you find the Empirical Formula when you only have Substances and the Mass of each present element

Answer 16

To find the Empirical Formula, use this table:

|———–|———-|———–|
| Element | Symbol 1 | Symbol 2 |
|———–|———-|———–|
| m (g) | | |
|———–|———-|———–|
| M (g/mol) | | |
|———–|———-|———–|
| n (mol) | | |
|———–|———-|———–|
| Simplest | | |
| Whole | | |
| Number | | |
| Ratio | | |
|———–|———-|———–|

(n = m/M)

Question 18

🟠 What Is The Law of Conservation of Mass in Chemical Reactions

Answer 18

Total Mass of Reactants = Total Mass of Products

Question 19

🟠 Find the Empirical Formula of these substances:

N: 82.4%
H: 17.6%

Answer 19

|———–|———-|———–|
| Element | Nitrogen | Hydrogen |
|———–|———-|———–|
| m (g) | 82.4 | 17.6 |
|———–|———-|———–|
| M (g/mol) | 14 | 1 |
|———–|———-|———–|
| n (mol) | 5.89 | 1 |
|———–|———-|———–|
| Simplest | | |
| Whole | 1 | 3 |
| Number | | |
| Ratio | | |
|———–|———-|———–|

So the ratio is 1 : 3, meaning the Emprical Formula is NH3

Question 20

🟠 Find the Empirical Formula of these substances:

C: 0.60g
H: 0.10g
O: 0.80g

Answer 20

|———–|———-|———–|———|
| Element | Carbon | Hydrogen | Oxygen |
|———–|———-|———–|———|
| m (g) | 0.60 | 0.10 | 0.80 |
|———–|———-|———–|———|
| M (g/mol) | 12 | 1 | 16 |
|———–|———-|———–|———|
| n (mol) | 0.05 | 0.1 | 0.05 |
|———–|———-|———–|———|
| Simplest | | | |
| Whole | (5) 1 | (10) 2 | (5) 1 |
| Number | | | |
| Ratio | | | |
|———–|———-|———–|———|

So the ratio is 1 : 2 : 1, meaning the Empirical formula is CH2O.

Question 21

🟢 How to find the Limiting Reactant example question:

ZnCO3 + 2HCl –> ZnCl2 + H2O + CO2
6.25g of ZnCO3 was added to a solution containing 1.825g if HCl

Which reactant is the limiting reactant?

(First state the method, then work out the answer)

Answer 21

You do Mass/Mr to get the number of mols
Then work the ratio between the two substances in question to work out the required amount of mols needed for each susbstance.
Then compared the required amount with the actual amount to get the answer.

6. 25g/125 = 0.05 mol (needs 2x0.05 = 0.1 mol HC.)
7. 825g/36.5 = 0.05 mol HC, (but 0.1 mol is needed)

This means ZnCo3 is in excess, and
HCl is the Limiting Reactant

Question 22

🟢 How to work out what unknown mass of one substance reacts with the known mass of another substance

(e.g What mass of Oxygen reacts with 12g of Magnesium? - 2Mg + O2 –> 2MgO)

Answer 22

Firstly you do the Mass/Mr to get the number of mols of the known substance (only include the small subscript numbets like Fe2 for examples, but not 2Fe)
Once you have the number of mols, find the ratio between the two kniw ans unknown substances. If it is 1:2 for example, times the number of mols by 2.
Then take this number and times it my the Mr (Mols x Mr) to get the Mass (when timsing with the Me, only include the small subscript numbets like Fe2 for examples, but not 2Fe)
Now, you have worked out your unknown mass.

n(Mg) = 12g / 24 = 0.5mol
2Mg : 1O
n(O2) = 0.5 / 2 = 0.25 mol
m(O2) = 0.25 x 32 = 8g

The mass of the Oxygen is 8g

Question 23

🟢 How to work out what unknown mass of one substance reacts with the known mass of another substance

(e.g Calculate the mass of aluminium that can be formed from 1020 g of aluminium oxide 2Al2O3 → 4Al + 3O2)

Answer 23

Firstly you do the Mass/Mr to get the number of mols of the known substance (only include the small subscript numbets like Fe2 for examples, but not 2Fe)
Once you have the number of mols, find the ratio between the two kniw ans unknown substances. If it is 1:2 for example, times the number of mols by 2.
Then take this number and times it my the Mr (Mols x Mr) to get the Mass (when timsing with the Me, only include the small subscript numbets like Fe2 for examples, but not 2Fe)
Now, you have worked out your unknown mass.

n(Al2O3) = 1020g / 102 = 10 mols
2Al2O3 : 4Al
n(Al) = 10 x 2 = 20 mols
m(O2) = 20 x 27 = 540g

The mass of the Aluminium is 540g

Question 24

🟢 How do you find the Empirical Formula when you only have Substances and the Perecentage of each present element

Answer 24

When you are given the percentage instead of tne mass, you do the exact same thing, as long as all of the percentages get a sum of 100%, because then everything is in proportion, which is what Empirical Formula needs to be.

Use this table:

|———–|———-|———–|
| Element | Symbol 1 | Symbol 2 |
|———–|———-|———–|
| m (g) | | |
|———–|———-|———–|
| M (g/mol) | | |
|———–|———-|———–|
| n (mol) | | |
|———–|———-|———–|
| Simplest | | |
| Whole | | |
| Number | | |
| Ratio | | |
|———–|———-|———–|

Question 25

🟢 What is a Limiting Reactant?

Answer 25

In a chemical reaction involving two reactants, it is common to use an excess of on eof the reactants. This ensures the hole of the other reactant is used, and the largest amount of product can be made.

The reactant that is completely used up is called the Limiting Reactant, because it limits the amount if products made.

Question 26

🟢 How do you find a Limiting Reactant? (e.g TiCl4 (1 mol) + 4Na (4 mol) –> Ti (1 mol) + 4NaCl (4 mol))

Answer 26

Firstly, you need to get your mole ratio, and then look at the amount if mols are present if each substance. Then, compare the mols of the reactants using the mole ratio (mentioned first), and find which one doesn’t have enough mols. You can then use the number of mols is actually present in the limiting reactant (not the required amount), and use the ratio to find out the number of mols that are in the products. You can always convert the mols into grams, by doing the Mols x Molar Mass = Mass (g).

What the Limiting Reactant?

The Equation is:
TiCl4 + 4Na –> Ti + 4NaCl

There are only:
4 mols of TiCl4 avaliable
and 4 mols of 4Na available

The Mole Ratio of the TiCl4 : 4Na : Ti : 4NaCl
1 : 4 : 1 : 4
-How Many mols of 4N are needed
- 4 mols of Ti x 4 (as the ratio is 1:4) = 16 mols of 4Na needed, but
there are only 4 mols of it avaliable, making it the limiting reactant
- 4 mols of 4Na / 4 (as the ratio is 1:4) = 1 mol of TiCl4 needed, and
there are 4 mols of it avaliable, makimg this reactant in excess.

When working out the mols of the products, you can use the amount of mols avaliable in the limiting reactant, and scale it to the products in line with the ratio

• 4 mols of 4Na / 4 = 1 mol of Ti made
• 4 mols of 4Na x 1 = 4 mols of 4NaCl made

(To get the mass in grams of these products, times the amount of mols by the substance’s molar mass. This gets:
1 x 48 = 48g
and
4 x 58.5 = 234g).

Question 27

🟠 What is the Yield of a Chemical Reaction and What is the Percentage Yield of a Chemical Reaction?

Answer 27

Yield:
The yield of a chemical reaction describes how much product is made.

Percentage Yield:
The Percentage Yield of a chemical reaction describes how much product is made compared with the maximum amount that could be made 100%

Question 28

🟠 Why might a reaction not give 100% Yield?

Answer 28

It could be:

• A reversible reaction: as products form they react to re-firm the reactants again. You show reversibke reactions using this symbol ⇌, instead of the normal arrow between the reactants and products. Chemists can amnipulste reversible reaction by the conditios they chooses in the reaction vessels in chemical plants
• Side reactions: some reactants may react to give unexpected or unwanted peoducts in alternative reactions
• Losses in separation: some of the desired product maybe lost in separation from the reaction mixture
• Losses in purification: the reactants mau not be pure (as in the case of lime kiln)

Question 29

🟠 Equation to work out the Percentage Yield

Answer 29

(Actual Mass of product formed / Maximum theoretical mass of products possible) x 100 = Percentage Yield

Question 30

🟠 How to Calculate the Percentage Yield

(e. g) Calculate the mass of iron that can be formed from 126g of iron oxide - Equation: Fe2O3 + 3CO –> 2Fe + 3CO2
b) 78.5g of Iron was actually formed in this reaction. Calculate the Percentage Yield)

Answer 30

First you do the Mass / Molar Mass to get the mols. Then you scale the mols in terms of the molar ratio between each substance. Next you get the mols that you now have, and times it by the Mr of the substance you are trying to find the mass of. This will give you the maximum theoretical mass of the substance. After this you will given the mass of the substance that is actually formed in the reaction, and from here you can just follow the equation: (Actual Mass of product formed / Maximum theoretical mass of products possible) x 100 = Percentage Yield.

a) 
n(Fe2O3) = 126 / 160 = 0.7875
Ratio: 1:2
0.7875 x 2 = 1.575
m(2Fe) = 1.575 x 56 = 88.2g (This is the Maximum theoretical mass possible)

b)
(78.5g / 88.2g) x 100 = 89.0%.

Question 31

🟠 What is Atom Economy? (+Why is it Important)

Answer 31

The atom economy that a reaction uses, is a measure of what percentage of the mass of reactants become useful products, and is caluclated using the balanced chemical equation for the reaction.

It is important for sustainable development and for economic reasons to use reactions with high atom economy, i.e, it is important to maximise atom economy in industrial processes to conserve the Earth’s resources and minimise pollution

Question 32

🟠 Equation to work out Percentage Atom Economy

Answer 32

Atom Economy = (Relative Formula Mass of Desired Product / Sum of Relative Formula Masses of All Reactants) x 100

Question 33

🟠 What can effect the decisions when deciding the best way to make certain products, besides using Atom Economy:

Answer 33

Chemical Companies use reactions to make products they sell. They need to consider:

• Percentage Yield
• Reaction Rate
• Reversibility (Equilibrium Position)
• Energy Cost
• Cost Of Raw Materials/Reactants
• Environmental Impact (Conserving Limited Resources, i.e, Fossil Fuels and Reducing Pollution)
• Waste Products (How Much Waste and How To Dispose Of It Safely

Question 34

🟠 How to Calculate Atom Economy (e.g Calculate the atom economy to form copper (II) oxide from copper (II) carbonate.

Equation:
WO3 + 3H2 –> W + 3H2O)

Answer 34

Firstly, calculate the total Mr of all the reactants. Then calculate the Mr of the desired products. After this, you do: (Mr of desired products / Mr of reactants) x 100 = to equal the atom economy of the substance in that equation.

CuCO3: 63.5 + 12 + (16x3) = 123.5
CuO: 63.5 + 16 = 79.5

(79.5 / 123.5) x 100 = 64%.

So, the atom economy to form copper (II) oxide from copper (II) carbonate is 64%.

Question 35

🟠 How is Volume changed or affected, if there are equal amounts, in moles, but for different gases

+ What is the volume of 1 mole of any gas at room temperature and room Pressure

Answer 35

No matter what, equal amounts, in moles, of any gas occupy the same volume under the same conditions, temperature, and pressure.

The volume of 1 mole of any gas at room temperature (20°C) and room pressure (1 atm) is: 24dm^3.

Question 36

🟠 Equation for the Amount of Gas and Volume

Answer 36

Number of moles of Gas (mol) = Volume of Gas (dm^3) / 24dm^3

Question 37

🟠 3 Steps on how to work out Questions Involving Reacting Masses, Gases, and Solutions.

Answer 37

Step 1: Convert the data given in tne question into amount, in moles

Step 2: Use the mole ratio of substances in the balanced chenical equation to calculate the amount, in moles, of the substance required.

Step 3: Convert the amount, in moles, back into the data required i.e, mass (g), or volume (dm^3), or concentration (mol/dm^3 or g/dm^3).

Question 38

🟠 Worked Examples:

1. Find the volume, measured at Room Temp, of 8 moles of O2 gas
2. A party balloon has an approximate volume of 12,500cm^3. Calculate the number of moles of helium. He used gas to fill the balloon.
3. Find the volume, measured at Room Temperature, of 4g of Methane CH4 gas

Answer 38

1. 8 x 24 = 192dm^3
2. 12,500cm^3 = 12.5dm^3
   
   12. 5dm^3 x 24 = 300ml
3. 4g/16 = 0.25mol
   V = 0.25mol x 24 = 6dm^3

Question 39

🟠 Molar Gas Volumes: Equations for Volume (dm^3), Mols (n), and for what 1dm^3 is equal to?

Answer 39

Volume (dm^3) = 24 x n (Mol)

n (Mol) = Volume (dm^3) / 24

1dm^3 = 1000cm^3

Question 40

🟠 How do you calculate the volume of one substance in dm^3, when you are only given the number of mols in another.

(Example: Calculate the volume of NH3 produced in cm^3 from the reaction of 6000cm^3 H2 with an excess of

Equation: N2N2 + 3H2 –> 2NH3)

Answer 40

Firstly, you can get volume inti dm^3. Then find the mole ratio of all the substances in the equation, which is 1 : 3 : 2. Then you use this ratio to find the proportional amount of molesn for each substance, and then times the moles but 24 to get the volume.

Example Question:

6000cm^3 = 6dm^3
6 / 24 = 0.25

3H2 has 0.25mols
So, 2NH3 has 0.16

0.16 x 24 = 4dm^3

Question 41

🟠 What is the Concentration of a Solution? (+ What is it measured in and the Equations to get to them)

Answer 41

The concentration, c, of a solution is the amount of substance per unit volume

It is usually measured in:

• mol/dm^3 Concentration = Amount / Volume
  c (mol/dm^3) = n (mol) / V (dm^3)

Or

-g/dm^3 Concentration = Amount / Volume
c (g/dm^3) = m (g) / V (dm^3)

Question 42

🟠 Example Question: Calculate the Concentration in mol/dm^3 of 2 moles of NaCl in 500cm^3

Answer 42

2 moles / 0.5 dm^3 = 4 mol/dm^3

Moles / Volume dm^3 = mol/dm^3

Question 43

🟠 Example Question: Calculate the Concentration in g/dm^3 of 2 moles of H2SO4 in 500cm^3

Answer 43

Molar Mass of H2SO4: 98

2 moles x 98 = 196g

196g x 0.5dm^3 = 392g/dm^3

(Moles x Molar Mass = Mass
Mass x Volume (dm^3) = g/dm^3)

Question 44

🟠 Calculate the number of moles in 25cm^3 of 1.5 mol/dm^3

Answer 44

1.5 mol/dm^3 x 0.025dm^3 = 0.0375 moles

Concentration x Volume
(mol/dm^3 x dm^3 = moles)

Question 45

🟠 Calculate the concentration of 0.100 mol/dm^3 of NaOH in g/dm^3

Answer 45

Molar Mass of NaOH: 40

0.100 x 40 = 4g/dm^3

(mol/dm^3 x Molar Mass = g/dm^3)

Question 46

🟠 How would you prepare the following solution from the solution given, and water:

30cm^3 of 1 mol/dm^3 from 3 mol/d ^3

Answer 46

30cm^3 x 1 mol/dm^3 = 30 mol

30 mol / 3 mol/dm^3 = 10cm^3 of Solution

Overall Solution and Water = 30cm^3

30cm^3 - 10cm^3 = 20cm^3 of Water

10cm^3 of Solution
20cm^3 of Water

(Volume (cm^3) x mol/dm^3 = mol
Moles / mol/dm^3 = Volume (cm^3) of Solution
Overall Solution and Water - Amount of Solution = Amount of Water)

Question 47

🟢 Titration Calculation: There is 25.0 cm^3 of a solution of sodium hydroxide solution required 21.5 cm^3 of 0.100 mol/dm^3 sulfuric acid for neutralisation.

H2SO4(aq) + 2 NaOH(aq) → Na2SO4(aq) + 2 H2O(l)

Find the concentration of the sodium hydroxide solution in mol/dm^3

Answer 47

cv = 0.1 x (21.5/1000) = 2.15x10^-3 moles
(Concentration x Volume)
Mole Ratio = 1:2, so 2.15x10^-3 x 2 = 4.3x10-3
Moles/Volume = 4.3x10-3 - (25/1000) = 0.172mol/dm^3

Ans = 0.172mol/dm^3

Question 48

🟢 Find the volume of 1.20 mol/dm3 hydrochloric acid that reacts with 25.0 cm3 of 1.50 mol/dm3 sodium hydroxide.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Answer 48

cv = 1.5x(25/1000) = 0.0375 moles
(Concentration x Volume)
Mole Ratio = 1:1, so 0.0375 x 1 = 0.0375
Moles/Concentration = 0.0375 / 1.2 = 0.03125dm^3

Ans = 0.03125dm^3 = 31.25cm^3

Question 49

🟠 How do you find uncertainty in an experiment

Answer 49

Whenever a measurement is made there is always some uncertainty about the result obtained.

This is often recorded after a measurement as a ± value (e.g 21 ± 2°C)

In an experiment that is repeated severaltimes, the mean value can be found (having omitted any anomalous results).

The uncertainty can then be estimated from the range of results used to find the mean.

Question 50

🟠 Worked Example: The volume if gas produced in a reaction was measured six times.

The values were: 83, 71, 78, 61, 82, 79 cm^3.

What is the mean value with uncertainty?

Answer 50

Anomalous result = 61cm^3

(71+78+79+82+83)/5 = 79cm^3

The range is from 71 to 83 cm^3 so the farthest range from 79cm^3 is from any of these numbers is 8cm^3 (7cm^39-71cm^3 = 8cm^3)

Therefore, the mean with uncertainty is: 79cm^3 ± 8cm^3

Question 51

🟠 Find any anomalous results in the data befire claculating the mean value with uncertainty fir the volume of acid needed to neutralise an alkali in a titration.

The results are: 24.2, 25.6, 24.4, 24.1 cm^3

Answer 51

Anomalous result = 25.6cm^3

(24.1+24.2+24.4)/3 = 24.23cm^3, but to 1dp = 24.2cm^3

The range is from 24.1 to 24.4, so the farthest range 24.2cm^3 is 0.2cm^3 (24.4cm^3-24.2cm^3 = 0.2cm^3)

Therefore the mean with uncertainty is 24.2cm^3 ± 0.2cm^3

Question 52

🟠 Equation for Concentration

Answer 52

Concentration (g/dm^3) = Amount of solute (g) / Volume of Soution

Question 53

🟢 Which solution goes in what piece of equipment in a Titration?

Answer 53

In a titration, you always have one solution with a concentration that you know accurately. This goes in the burette. Then you can place the other so,ution, with an unknown concentration, in a concial flash. This is done using a volumetric pipette. This ensures you know the volumeof this solution accurately. This result from the titration is used to calculate the number of moles of thesubstsnce dissolved in the solution in the concial flask.